What Is the Probability of Hearing Counting Stars During a 30-Minute Drive?

[mjordan2nd]

This is not a homework question, but something I am curious about because of a debate with my friend.

Lets say between three radio stations the song Counting Stars plays 12 times between the hours of 10am and 6pm. The song has an approximate run time of 4:20. What is the probability I will hear that song on one of those three stations in a 30 minute drive?

I tried solving the problem using the following logic. In those hours, that song played on the radio for a total of 51.96 minutes, or 10.8% of the time (51.96/480=.108). Therefore the probability of not hearing that song at any given time is 89.2%. A 30 minute time period can be divided into ~7 4.33 minute periods so the probability of not hearing that song in that time period is .892^7~.45. Therefore the probability of hearing that song on a 30 minute drive is 55%.

I have a feeling that this logic is incorrect, though. Can anyone help me out?

Thanks.

 

[haruspex]

Something I'm missing - are you listening to all 3 stations at once, randomly flipping between them at a great rate, or listening to the same one all 30 minutes?
Your 10.8% analysis assumes it's never playing on two stations at once.
The next calculation doesn't take into account that the song can start any second, not just aligned with the 4.33 minute periods you chopped the 30 minutes into.

 

[mjordan2nd]

Lets say I'm flipping between them at a great rate.

You've pointed out two flaws in my argument, however I'm not sure how to correct them.

 

[Ray Vickson]

This is not a homework question, but something I am curious about because of a debate with my friend.

Lets say between three radio stations the song Counting Stars plays 12 times between the hours of 10am and 6pm. The song has an approximate run time of 4:20. What is the probability I will hear that song on one of those three stations in a 30 minute drive?

I tried solving the problem using the following logic. In those hours, that song played on the radio for a total of 51.96 minutes, or 10.8% of the time (51.96/480=.108). Therefore the probability of not hearing that song at any given time is 89.2%. A 30 minute time period can be divided into ~7 4.33 minute periods so the probability of not hearing that song in that time period is .892^7~.45. Therefore the probability of hearing that song on a 30 minute drive is 55%.

I have a feeling that this logic is incorrect, though. Can anyone help me out?

Thanks.

Because you switch rapidly between the stations, you will hear the song if it is playing at all. So, you are asking for the probability that at least part of at least one of the songs overlaps a given 30-minute interval, given that the song is played 12 times during the long time interval, with random start times and no more than three overlaps at any time. If we drop the last requirement, and also make some pretty sweeping assumptions we can get a solution; this will be an approximate solution to the original problem because it allows for more than three playings simultaneously, while the original problem does not. However, given that such overlap of 4 or more will be pretty improbable, the approximation ought to be reasonable. Basically, we are replacing the problem by one having 12 radio stations, and are ignoring the possibility that 4 or more stations are playing the song at the same time. We will also assume independent start-times of the 12 playings, but that is not really true, either. For example, if Station 1 is playing the song at some time t, it cannot play it again before the current playing is finished, but we shall ignore that reality. Finally: stations break for news and weather, but we are ignoring that as well. (You could adjust the model to account for that; it gets a bit more complicated, but remains "doable".)

You do not say whether the whole song must run in the long interval (or whether, for example, the interval can start or end partway through one of the playings). I will assume whole playings only, so the start-time of each playing must be in the interval from 10am to (6pm - (4min + 20 sec)). If we assume independent, uniformly-distributed start-times, we want the probability that at least one start-time falls in a given interval of length (30min + 4 min + 30 sec). That will be the probability of >= 1 success in a binomial distribution for 12 trials with success probability p = (30 min + 4 min + 30 sec)/(8 hr - 4 min - 30 sec) per trial. To compute p we must convert the numerator and denominator to some common time unit.

In general, some fairly simple-sounding probability problems can be notoriously hard to solve exactly, and this seems like one of those. If you want a better analysis, it might be necessary to resort to simulation/Monte-Carlo methods.